Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $13.1$ years; the standard deviation is $2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living between $17.1$ and $19.1$ years.
$13.1$ $11.1$ $15.1$ $9.1$ $17.1$ $7.1$ $19.1$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $13.1$ years. We know the standard deviation is $2$ years, so one standard deviation below the mean is $11.1$ years and one standard deviation above the mean is $15.1$ years. Two standard deviations below the mean is $9.1$ years and two standard deviations above the mean is $17.1$ years. Three standard deviations below the mean is $7.1$ years and three standard deviations above the mean is $19.1$ years. We are interested in the probability of a lion living between $17.1$ and $19.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the lions will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the lions will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of lions between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular lion living between $17.1$ and $19.1$ years is $\color{orange}{2.35\%}$.